Answer:
The function f(x) is positive in the interval (-≠,-2.5) ∪ (1,∞)
Step-by-step explanation:
we have
[tex]f(x)=2x^{2}+3x-5[/tex]
This is a vertical parabola open upward (the leading coefficient is positive)
The vertex is a minimum
The coordinates of the vertex is the point (h,k)
step 1
Find the vertex of the quadratic function
Factor the leading coefficient 2
[tex]f(x)=2(x^{2}+\frac{3}{2}x)-5[/tex]
Complete the square
[tex]f(x)=2(x^{2}+\frac{3}{2}x+\frac{9}{16})-5-\frac{9}{8}[/tex]
[tex]f(x)=2(x^{2}+\frac{3}{2}x+\frac{9}{16})-\frac{49}{8}[/tex]
Rewrite as perfect squares
[tex]f(x)=2(x+\frac{3}{4})^{2}-\frac{49}{8}[/tex]
The vertex is the point (-\frac{3}{4},-\frac{49}{8})
step 2
Find the x-intercepts (values of x when the value of f(x) is equal to zero)
For f(x)=0
[tex]2(x+\frac{3}{4})^{2}-\frac{49}{8}=0[/tex]
[tex]2(x+\frac{3}{4})^{2}=\frac{49}{8}[/tex]
[tex](x+\frac{3}{4})^{2}=\frac{49}{16}[/tex]
take the square root both sides
[tex]x+\frac{3}{4}=\pm\frac{7}{4}[/tex]
[tex]x=-\frac{3}{4}\pm\frac{7}{4}[/tex]
[tex]x_1=-\frac{3}{4}+\frac{7}{4}=1[/tex]
[tex]x_2=-\frac{3}{4}-\frac{7}{4}=-2.5[/tex]
therefore
The function f(x) is negative in the interval (-2.5,1)
The function f(x) is positive in the interval (-≠,-2.5) ∪ (1,∞)
see the attached figure to better understand the problem
